Q6 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Chemistry

The temperature of 4.0 moles of a gas occupying $5 \mathrm{dm}^3$ at $3.32 \mathrm{bar}$ is $\left(R=0.083 \mathrm{bar} \mathrm{dm}^3 \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$

  • A
    25 K
  • Correct answer
    B
    50 K
  • C
    75 K
  • D
    100 K

Solution

The amount of the gas, $n=4.0 \mathrm{~mol}$ Volume of gas $Y=5 \mathrm{dm}^3$ Pressure of the gas $p=3.32$ bar $$ R=0.083 \text { bar dm }^3 \mathrm{~K}^{-1} \mathrm{~mol}^{-1} $$ From the ideal gas equation, we get $$ p V=n R T \Rightarrow T=\frac{p V}{n R}=(3 \cdot 32 \times 5) /(4 \times 0 \cdot 083) $$ Hence, required temperature is $=50 \mathrm{~K}$

Hence, the correct answer is option: 'B'.

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